Re: question on UGB
avinash said:
from gregorian i read that for SC circuits the UGB(w0) of opamp should be greater than 1/Tsh.
w0 > 15/Tsh
Tsh is the charging discharging time of load.
Tsh=1/(2f),where f is clock frequency.
now my question is if my clock is 1Mhz,w0 comes to be 30Mhz.is it not sounding strange.
please help
it depends on ur spec. of accuracy ( or resolution).
as u may know, the step response of a single pole (ωp) SC circuit may be represented as : Vout(t)=Vi(1-exp(-t/α)), where α=1/[(1+Aβ)ωp)] is the time constant of the closed loop system. So, the difference voltage (Δ) between Vout(t1) and Vout(0) is Δ=Vi*exp(-t1/α)).
==> t1=-α×ln(Δ/Vi) = -ln(Δ/Vi)/[(1+Aβ)ωp)] ≈ ln(Vi/Δ)/(Aβωp)=ln(Vi/Δ)/(βω0)
that is the difference will decrease as t1 increase
==> t1 >= 1/(2f) to achieve the desired difference voltage.
==> 1/(2f) >= ln(Vi/Δ)/(βω0)
==> ω0 >= 2f*ln(Vi/Δ)/β
==> 2*pi*f0 >= 2f*ln(Vi/Δ)/β ==> f0 >=f*ln(Vi/Δ)/(β*pi)
if ur spec. is Vi/Δ >= 1000, β=1 ==> ω0 >= 13.8f ==> f0 >=4.39f
Vi/Δ >= 1000, β=1/2 ==> ω0 >= 27.6f ==> f0 >=8.78f
Vi/Δ >= 10000, β=1 ==> ω0 >= 18.4f ==> f0 >=5.86f
so if f=1MHz, in order to achieve accuracy of 0.1% (i.e. Vi/Δ >= 1000) ,u need to design ur f0 >=4.39MHz for β=1.
the above derivation show that u don't need such high BW of ur UGB for 1MHz sampling clock, but always keep in mind ,there exist variation of process , so u should always choose large value of f0.
hopes this help u.